After having dealt with the basic properties of elliptic functions in the previous post we shall now focus on the addition formulas for them. These are used to express the functions of sum of two arguments in terms of functions of each argument separately. The additions formulas are algebraic in nature and in fact, in general any function with an algebraic addition formula is necessarily an elliptic function or a limiting case of it. We will not prove this general result here as it requires the use of theory of functions, but we shall be content to derive the formulas for the specific elliptic function which we are considering here.

Let us assume that the variables $u_{1}$ and $u_{2}$ vary in such a way that their sum $u_{1} + u_{2}$ is a constant. And we set: \begin{align} s_{1} = \text{sn}(u_{1}, k),\, c_{1} = \text{cn}&(u_{1}, k),\, d_{1} = \text{dn}(u_{1}, k)\notag\\ s_{2} = \text{sn}(u_{2}, k),\, c_{2} = \text{cn}&(u_{2}, k),\, d_{2} = \text{dn}(u_{2}, k)\notag\\ u_{1} + u_{2} &= a\notag \end{align} so that $$\frac{du_{2}}{du_{1}} = -1$$ Then we have \begin{align} \frac{d}{du_{1}}(s_{1} + s_{2}) &= c_{1}d_{1} - c_{2}d_{2}\notag\\ \frac{d}{du_{1}}(s_{1}c_{2} + s_{2}c_{1}) &= s_{1}s_{2}d_{2} + c_{1}d_{1}c_{2} - s_{2}s_{1}d_{1} - c_{2}d_{2}c_{1}\notag\\ &= (d_{1} - d_{2})(c_{1}c_{2} - s_{1}s_{2})\notag\\ \frac{d}{du_{1}}(d_{1} + d_{2}) &= -k^{2}(s_{1}c_{1} - s_{2}c_{2})\notag\\ &= -k^{2}s_{1}c_{1}(c_{2}^{2} + s_{2}^{2}) + k^{2}s_{2}c_{2}(c_{1}^{2} + s_{1}^{2})\notag\\ &= -k^{2}(c_{1}c_{2} - s_{1}s_{2})(s_{1}c_{2} - s_{2}c_{1})\notag \end{align} Since we have \begin{align} -k^{2}(s_{1}^{2}c_{2}^{2} - s_{2}^{2}c_{1}^{2}) &= -k^{2}(s_{1}^{2}(1 - s_{2}^{2}) - s_{2}^{2}(1 - s_{1}^{2}))\notag\\ &= -k^{2}(s_{1}^{2} - s_{2}^{2})\notag\\ &= d_{1}^{2} - d_{2}^{2}\notag \end{align} It therefore follows that $$(d_{1} + d_{2})\frac{d}{du_{1}}(s_{1}c_{2} + s_{2}c_{1}) = (s_{1}c_{2} + s_{2}c_{1})\frac{d}{du_{1}}(d_{1} + d_{2})$$ From the above it is clear that the ratio $(s_{1}c_{2} + s_{2}c_{1})/(d_{1} + d_{2})$ is a constant independent of $u_{1}$ as long as the sum $u_{1} + u_{2} = a$ is a constant. Therefore value of this ratio is obtained by putting $u_{1} = 0$ and $u_{2}= a$ and we have $$\frac{s_{1}c_{2} + s_{2}c_{1}}{d_{1} + d_{2}} = \frac{\text{sn}(a, k)}{1 + \text{dn}(a, k)} = \frac{\text{sn}(u_{1} + u_{2}, k)}{1 + \text{dn}(u_{1} + u_{2}, k)}$$ Again we can see that \begin{align} \frac{s_{1}c_{2} - s_{2}c_{1}}{d_{1} - d_{2}} &= -\frac{1}{k^{2}}\cdot \frac{d_{1} + d_{2}}{s_{1}c_{2} + s_{2}c_{1}} = \text{ a constant independent of }u_{1}\notag\\ &= \frac{\text{sn}(a, k)}{\text{dn}(a, k) - 1}\text{ by putting } u_{1} = a, u_{2} = 0\notag\\ &= \frac{\text{sn}(u_{1} + u_{2}, k)}{\text{dn}(u_{1} + u_{2}, k) - 1}\notag \end{align} We therefore have the following results: \begin{align} \frac{\text{dn}(u_{1} + u_{2}, k) + 1}{\text{sn}(u_{1} + u_{2}, k)} &= \frac{d_{1} + d_{2}}{s_{1}c_{2} + s_{2}c_{1}}\tag{A}\\ \frac{\text{dn}(u_{1} + u_{2}, k) - 1}{\text{sn}(u_{1} + u_{2}, k)} &= \frac{d_{1} - d_{2}}{s_{1}c_{2} - s_{2}c_{1}}\tag{B} \end{align} Subtracting the above equations we get $$\frac{2}{\text{sn}(u_{1} + u_{2}, k)} = \frac{d_{1} + d_{2}}{s_{1}c_{2} + s_{2}c_{1}} - \frac{d_{1} - d_{2}}{s_{1}c_{2} - s_{2}c_{1}} = \frac{2(s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1})}{s_{1}^{2}c_{2}^{2} - s_{2}^{2}c_{1}^{2}}$$ and therefore $$\text{sn}(u_{1} + u_{2}, k) = \frac{s_{1}^{2} - s_{2}^{2}}{s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1}}$$ Adding the equations $(A)$ and $(B)$ we get $$2\text{ds}(u_{1} + u_{2}, k) = \frac{d_{1} + d_{2}}{s_{1}c_{2} + s_{2}c_{1}} + \frac{d_{1} - d_{2}}{s_{1}c_{2} - s_{2}c_{1}} = \frac{2(s_{1}c_{2}d_{1} - s_{2}c_{1}d_{2})}{s_{1}^{2} - s_{2}^{2}}$$ and therefore $$\text{dn}(u_{1} + u_{2}, k) = \frac{s_{1}c_{2}d_{1} - s_{2}c_{1}d_{2}}{s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1}}$$ In similar fashion one can obtain the following formulas: \begin{align} \frac{\text{cn}(u_{1} + u_{2}, k) + 1}{\text{sn}(u_{1} + u_{2}, k)} &= \frac{c_{1} + c_{2}}{s_{1}d_{2} + s_{2}d_{1}}\notag\\ \frac{\text{cn}(u_{1} + u_{2}, k) - 1}{\text{sn}(u_{1} + u_{2}, k)} &= \frac{c_{1} - c_{2}}{s_{1}d_{2} - s_{2}d_{1}}\notag \end{align} Adding the above equations we get $$2\text{cs}(u_{1} + u_{2}, k) = \frac{2(s_{1}c_{1}d_{2} - s_{2}c_{2}d_{1})}{s_{1}^{2} - s_{2}^{2}}$$ and finally we have $$\text{cn}(u_{1} + u_{2}, k) = \frac{s_{1}c_{1}d_{2} - s_{2}c_{2}d_{1}}{s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1}}$$ To summarize we have the following addition formulas: \begin{align} \text{sn}(u_{1} + u_{2}, k) &= \frac{s_{1}^{2} - s_{2}^{2}}{s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1}}\tag{1}\\ \text{cn}(u_{1} + u_{2}, k) &= \frac{s_{1}c_{1}d_{2} - s_{2}c_{2}d_{1}}{s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1}}\tag{2}\\ \text{dn}(u_{1} + u_{2}, k) &= \frac{s_{1}c_{2}d_{1} - s_{2}c_{1}d_{2}}{s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1}}\tag{3} \end{align} The above forms are hard to remember and don't look similar to addition formulas for circular functions. However they can be transformed by algebraic manipulations into a form which is similar to those of circular functions. This we will do in the next post. For the time being we can use these results to obtain expressions for $\text{sn}(u + K, k),\, \text{cn}(u + K, k),\, \text{dn}(u + K, k)$.
Putting $u_{1} = u$ and $u_{2} = K$ we get \begin{align} \text{sn}(u + K, k) &= \frac{\text{cn}(u, k)}{\text{dn}(u, k)} = \text{cd}(u, k)\tag{4}\\ \text{cn}(u + K, k) &= \frac{-k^{\prime}\,\text{sn}(u, k)}{\text{dn}(u, k)} = -k^{\prime}\,\text{sd}(u, k)\tag{5}\\ \text{dn}(u + K, k) &= \frac{k^{\prime}}{\text{dn}(u, k)} = k^{\prime}\,\text{nd}(u, k)\tag{6} \end{align} We can now put $u_{1} + K$ in place of $u_{1}$ in the addition formulas to get \begin{align} \text{sn}(u_{1} + K + u_{2}, k) &= \dfrac{\dfrac{c_{1}^{2}}{d_{1}^{2}} - s_{2}^{2}}{\dfrac{c_{1}c_{2}d_{2}}{d_{1}} + \dfrac{k'^{2}s_{1}s_{2}}{d_{1}^{2}}}\notag\\ \Rightarrow \frac{\text{cn}(u_{1} + u_{2}, k)}{\text{dn}(u_{1} + u_{2}, k)} &= \frac{c_{1}^{2} - d_{1}^{2}s_{2}^{2}}{c_{1}c_{2}d_{1}d_{2} + k'^{2}s_{1}s_{2}}\notag \end{align} and \begin{align} \text{dn}(u_{1} + K + u_{2}, k) &= \dfrac{\dfrac{k'c_{1}c_{2}}{d_{1}^{2}} + k'\dfrac{s_{1}s_{2}d_{2}}{d_{1}}}{\dfrac{c_{1}c_{2}d_{2}}{d_{1}} + \dfrac{k'^{2}s_{1}s_{2}}{d_{1}^{2}}}\notag\\ \Rightarrow \frac{k'}{\text{dn}(u_{1} + u_{2}, k)} &= \frac{k'c_{1}c_{2} + k's_{1}s_{2}d_{1}d_{2}}{c_{1}c_{2}d_{1}d_{2} + k'^{2}s_{1}s_{2}}\notag\\ \Rightarrow \text{dn}(u_{1} + u_{2}, k) &= \frac{c_{1}c_{2}d_{1}d_{2} + k'^{2}s_{1}s_{2}}{c_{1}c_{2} + s_{1}s_{2}d_{1}d_{2}}\notag\\ \Rightarrow \text{cn}(u_{1} + u_{2}, k) &= \frac{c_{1}^{2} - d_{1}^{2}s_{2}^{2}}{c_{1}c_{2} + s_{1}s_{2}d_{1}d_{2}} = \frac{1 - s_{1}^{2} - s_{2}^{2} + k^{2}s_{1}^{2}s_{2}^{2}}{c_{1}c_{2} + s_{1}s_{2}d_{1}d_{2}}\notag \end{align} Again \begin{align} \text{cn}(u_{1} + K + u_{2}, k) &= \dfrac{-\dfrac{k's_{1}c_{1}d_{2}}{d_{1}^{2}} - \dfrac{k's_{2}c_{2}}{d_{1}}}{\dfrac{c_{1}c_{2}d_{2}}{d_{1}} + \dfrac{k'^{2}s_{1}s_{2}}{d_{1}^{2}}}\notag\\ \Rightarrow \frac{\text{sn}(u_{1} + u_{2}, k)}{\text{dn}(u_{1} + u_{2}, k)} &= \frac{s_{1}c_{1}d_{2} + s_{2}c_{2}d_{1}}{c_{1}c_{2}d_{1}d_{2} + k'^{2}s_{1}s_{2}}\notag \end{align} Thus we can now write the addition formulas in a different form: \begin{align} \text{sn}(u_{1} + u_{2}, k) &= \frac{s_{1}c_{1}d_{2} + s_{2}c_{2}d_{1}}{c_{1}c_{2} + s_{1}s_{2}d_{1}d_{2}}\tag{7}\\ \text{cn}(u_{1} + u_{2}, k) &= \frac{1 - s_{1}^{2} - s_{2}^{2} + k^{2}s_{1}^{2}s_{2}^{2}}{c_{1}c_{2} + s_{1}s_{2}d_{1}d_{2}}\tag{8}\\ \text{dn}(u_{1} + u_{2}, k) &= \frac{c_{1}c_{2}d_{1}d_{2} + k'^{2}s_{1}s_{2}}{c_{1}c_{2} + s_{1}s_{2}d_{1}d_{2}}\tag{9} \end{align} We have so far developed the addition formulas in two different forms and we note that in each of the forms the formulas have a common denominator. Using them we have also obtained the expressions for $\text{sn}(u + K, k)$ etc.